Pearson Education Limited Edinburgh Gate Harlow Essex CM0 JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsoned.co.uk Pearson Education Limited 0 All rights reserved. No part of this publication ma be reproduced, stored in a retrieval sstem, or transmitted in an form or b an means, electronic, mechanical, photocoping, recording or otherwise, without either the prior written permission of the publisher or a licence permitting restricted coping in the United Kingdom issued b the Copright Licensing Agenc Ltd, Saffron House, 6 0 Kirb Street, London ECN 8TS. All trademarks used herein are the propert of their respective owners. The use of an trademark in this text does not vest in the author or publisher an trademark ownership rights in such trademarks, nor does the use of such trademarks impl an affiliation with or endorsement of this book b such owners. ISBN 0: -9-09-7 ISBN : 978--9-09- British Librar Cataloguing-in-Publication Data A catalogue record for this book is available from the British Librar Printed in the United States of America
60 CHAPTER Sstems of Equations x 0 Solve b the elimination method... x + = 7, -x + = -a + b = 0, a - b = - x 6 (q, a) (, ) x x x x Since A, B checks, it is the solution. We can also see this in the graph shown at left. Do Exercises and. In order to eliminate a variable, we sometimes use the multiplication principle to multipl one or both of the equations b a particular number before adding. EXAMPLE Solve this sstem: () () If we add directl, we get x + 9 = 7, and we have not eliminated a variable. However, note that if the in equation () were -6, we could eliminate. Thus we multipl b - on both sides of equation () and add: Then x + =, x + 6 =. -6x - 6 = -0 x + 6 = -x + 0 = -8 -x = -8 x =. # + 6 = + 6 = 6 = 8 =. Multipling b - on both sides of equation () Equation () Adding Solving for x Substituting for x in equation () Solving for We obtain,, or x =, =. This checks, so it is the solution. We can also see this in the graph at left. Do Exercise.. Solve b the elimination method: + x =, - + x = -. Sometimes we must multipl twice in order to make two terms opposites. EXAMPLE x + = 7, x + 7 = 9. Solve this sstem: () () We must first multipl in order to make one pair of terms with the same variable opposites. We decide to do this with the x-terms in each equation. We multipl equation () b and equation () b -. Then we get 0x and -0x, which are opposites. From equation : 0x + = 8 From equation : - 0x - = - 8 0 + = 7 = 7 Multipling b Multipling b - Adding Solving for Answers.,. a.,, b 66
. Solving b Elimination 6 Then x + # 7 = 7 x + 8 = 7 x = -6 x = -. We check the ordered pair -, 7. Substituting 7 for in equation () Solving for x Check: x + = 7 - + 7? 7-6 + 8 7 TRUE x + 7 = 9 - + 77? 9-60 + 89 9 We obtain -, 7, or x = -, = 7, as the solution. TRUE Do Exercises and. Solve b the elimination method.. x + = -8, 7x + 9 =. x - = 8, 7x - 8 = - Some sstems have no solution, as we saw graphicall in Section. and algebraicall in Example of Section.. How do we recognize such sstems if we are solving using elimination? EXAMPLE + x =, + x = -. Solve this sstem: () () If we find the slope intercept equations for this sstem, we get = -x +, = -x -. The graphs are parallel lines. The sstem has no solution. x x x Let s see what happens if we attempt to solve the sstem b the elimination method. We multipl b - on both sides of equation () and add: + x = - - x = 0 = 7. Equation () Multipling equation () b Adding, we obtain a false equation. The x-terms and the -terms are eliminated and we have a false equation. Thus, if we obtain a false equation, such as 0 = 7, when solving algebraicall, we know that the sstem has no solution. The sstem is inconsistent, and the equations are independent. - Do Exercise 6. 6. Solve b the elimination method: Answers + x =, + x = -.. -7, 00. -8, -8 6. No solution 67
6 CHAPTER Sstems of Equations Some sstems have infinitel man solutions. How can we recognize such a situation when we are solving sstems using an algebraic method? 8x x x 6 EXAMPLE Solve this sstem: - x = 6, - + 8x = -. () () We see from the figure at left that the graphs are the same line. The sstem has an infinite number of solutions. Suppose we tr to solve this sstem b the elimination method: - 8x = - + 8x = - 0 = 0. Multipling equation () b Equation () Adding, we obtain a true equation. We have eliminated both variables, and what remains is a true equation, 0 = 0. It can be expressed as 0 # x + 0 # = 0, and is true for all numbers x and. If an ordered pair is a solution of one of the original equations, then it will be a solution of the other. The sstem has an infinite number of solutions. The sstem is consistent, and the equations are dependent. SPECIAL CASES When solving a sstem of two linear equations in two variables: 7. Solve b the elimination method: x - = 0, -6x + = -0. 8. Clear the decimals. Then solve. (Hint: Multipl the first equation b 00 and the second one b 0.) 9. Clear the fractions. Then solve. Answers 0.0x + 0.0 = 0.0, 0.x - 0. = 0.7 x + =, x - = 7. Infinitel man solutions 8. x + =, x - = 7;, - 9. 9x + 0 =, 9x - = ; a 6, 7 b. If a false equation is obtained, such as 0 = 7, then the sstem has no solution. The sstem is inconsistent, and the equations are independent.. If a true equation is obtained, such as 0 = 0, then the sstem has an infinite number of solutions. The sstem is consistent, and the equations are dependent. Do Exercise 7. When solving a sstem of equations using the elimination method, it helps to first write the equations in the form Ax + B = C. When decimals or fractions occur, it also helps to clear before solving. EXAMPLE 6 Solve this sstem: 0.x + 0. =.7, 7 x + = 9. We have 0.x + 0. =.7, Multipling b 0 to clear decimals x + = 7, 7 x + = 9 Multipling b x + 7 = 9. to clear fractions We multiplied b 0 to clear the decimals. Multiplication b, the least common denominator, clears the fractions. The problem is now identical to Example. The solution is -, 7, or x = -, = 7. Do Exercises 8 and 9. 68
. Solving b Elimination 6 To use the elimination method to solve sstems of two equations:. Write both equations in the form Ax + B = C.. Clear an decimals or fractions.. Choose a variable to eliminate.. Make the chosen variable s terms opposites b multipling one or both equations b appropriate numbers if necessar.. Eliminate a variable b adding the respective sides of the equations and then solve for the remaining variable. 6. Substitute in either of the original equations to find the value of the other variable. Comparing Methods When deciding which method to use, consider this table and directions from our instructor. The situation is analogous to having a piece of wood to cut and three different tpes of saws available. Although all three saws can cut the wood, the best choice depends on the particular piece of wood, the tpe of cut being made, and our level of skill with each saw. METHOD STRENGTHS WEAKNESSES Graphical Can see solutions. Inexact when solutions involve numbers that are not integers. Solutions ma not appear on the part of the graph drawn. Substitution Elimination Yields exact solutions. Convenient to use when a variable has a coefficient of. Yields exact solutions. Convenient to use when no variable has a coefficient of. The preferred method for sstems of or more equations in or more variables. (See Section..) Can introduce extensive computations with fractions. Cannot see solutions quickl. Cannot see solutions quickl. 69
6 CHAPTER Sstems of Equations b Solving Applied Problems Using Elimination Let s now solve an applied problem using the elimination method. (We will solve man more problems in Section., which is devoted entirel to applied problems.) STUDY TIPS FIVE STEPS FOR PROBLEM SOLVING Remember to use the five steps for problem solving.. Familiarize ourself with the situation. Carefull read and reread the problem; draw a diagram, if appropriate; determine whether there is a formula that applies; assign letter(s), or variable(s), to the unknown(s).. Translate the problem to an equation, an inequalit, or a sstem of equations using the variable(s) assigned.. Solve the equation, inequalit, or sstem of equations.. Check the answer in the original wording of the problem.. State the answer clearl with appropriate units. 0. Bonuses. Monica gave each of the full-time emploees in her small business a ear-end bonus of $00 while each part-time emploee received $0. She gave a total of $000 in bonuses to her 0 emploees. How man full-time emploees and how man part-time emploees did Monica have? Answer 0. Full-time: 6; part-time: EXAMPLE 7 Stimulating the Hometown Econom. To stimulate the econom in his town of Brewton, Alabama, in 009, Dann Cottrell, co-owner of The Medical Center Pharmac, gave each of his full-time emploees $700 and each part-time emploee $00. He asked that each person donate % to a charit of his or her choice and spend the rest locall. The mone was paid in $ bills, a rarel used currenc, so that the business communit could easil see how the mone circulated. Cottrell gave awa a total of $6,000 to his emploees. How man full-time emploees and how man part-time emploees were there? Source: The Press-Register, March, 009. Familiarize. We let f = the number of full-time emploees and p = the number of part-time emploees. Each full-time emploee received $700, so a total of 700f was paid to them. Similarl, the part-time emploees received a total of 00p. Thus a total of 700f + 00p was given awa.. Translate. We translate to two equations. Total amount given awa is $6,000. Total number of emploees is. We now have a sstem of equations: 700f + 00p = 6,000, f + () (). Solve. First, we multipl b -00 on both sides of equation () and add: 700f + 00p = 6,000-00f - 00p = -700 00f = 8800 f =. Equation () Multipling b -00 on both sides of equation () Adding Solving for f Next, we substitute for f in equation () and solve for p: + p = p =.. Check. If there are full-time emploees and part-time emploees, there is a total of +, or, emploees. The full-time emploees received a total of $700 #, or $,00, and the part-time emploees received a total of $00 #, or $600. Then a total of $,00 + $600, or $6,000, was given awa. The numbers check in the original problem.. State. There were full-time emploees and part-time emploees. Do Exercise 0. 700f + 00p f + p p =. = = 6,000 70